Trig Problem: Where Did 50 Degrees Come From in Sin(2x)=.98?

In summary, the lecturer claims that the maximum for sin(2x) is located at x=45^o. However, this is only valid for the graph y=\sin{x} since the line of symmetry (the line that passes through the maximum) is located at x=90^o.
  • #1
xtrubambinoxpr
87
0

Homework Statement



Sin(2x)=.98

Homework Equations



I know you take arcsin of both sides and divide by 2 to get the answer which is 39, but where does this guy get 50 from? I would have assumed it occurred again at 2*39.. not 50. it is an online lecture.

The Attempt at a Solution



Arcsin of both sides solving give me 39 degrees
 

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  • #2
xtrubambinoxpr said:
I know you take arcsin of both sides and divide by 2 to get the answer which is 39, but where does this guy get 50 from? I would have assumed it occurred again at 2*39.. not 50. it is an online lecture.
Wrong assumption. Look at a graph of the sine function. Draw a line across just below y=1. You should see pairs of intercepts close together. How are they related?
 
  • #3
right it crosses the graph twice. just before the peak like he drew it, but how is that figured out? i know it is between half a period, or (pi) because it repeats itself after that. and in radians its .68.. help me figure this out without telling me! lol i really want to deduce it on my own
 
  • #4
With the graph [itex]y=\sin(x)[/itex] for [itex]0<x<\pi[/itex] what is the line of symmetry?
 
  • #5
Mentallic said:
With the graph [itex]y=\sin(x)[/itex] for [itex]0<x<\pi[/itex] what is the line of symmetry?

what do you mean by line of symmetry ? its in respect to the origin
 
  • #6
Do you know what a line of symmetry is? It'll just take a second to understand if you look it up, it's a simple concept. And no, it's not the origin (the origin is also not a line, it's a point).
 
  • #7
Mentallic said:
Do you know what a line of symmetry is? It'll just take a second to understand if you look it up, it's a simple concept. And no, it's not the origin (the origin is also not a line, it's a point).

well you mean that between 0 to pi you get sin of a value twice. like sin(60)=sin(120)..?
 
  • #8
xtrubambinoxpr said:
well you mean that between 0 to pi you get sin of a value twice. like sin(60)=sin(120)..?

Yes. Basically, the line of symmetry is at [itex]x=90^o[/itex] which means that the sine of an angle 2 degrees less than 90 is the same as the sine of the angle 2 degrees more, etc. In general, [itex]\sin(90^o+x)=\sin(90^o-x)[/itex].

Now, at what x value does [itex]y=\sin(2x)[/itex] have a maximum value? This is also the same as the line of symmetry problem I asked earlier.
 
  • #9
Mentallic said:
Yes. Basically, the line of symmetry is at [itex]x=90^o[/itex] which means that the sine of an angle 2 degrees less than 90 is the same as the sine of the angle 2 degrees more, etc. In general, [itex]\sin(90^o+x)=\sin(90^o-x)[/itex].

Now, at what x value does [itex]y=\sin(2x)[/itex] have a maximum value? This is also the same as the line of symmetry problem I asked earlier.

well it would be where sinx = 1 that's as high as it can go so x = 1/2
 
  • #10
xtrubambinoxpr said:
well it would be where sinx = 1 that's as high as it can go

Right!

xtrubambinoxpr said:
so x = 1/2

Umm... Think about this again.

Also, remember that if you're referring to degrees, you should add the sup tags (which is the x2 button located just above your post if you pick advanced edit) and put an 'o' into represent degrees.
 
  • #11
Mentallic said:
Right!



Umm... Think about this again.

Also, remember that if you're referring to degrees, you should add the sup tags (which is the x2 button located just above your post if you pick advanced edit) and put an 'o' into represent degrees.

well its at ∏/2 or 90°..
 
  • #12
Nope. if [itex]x=\pi / 2[/itex] then [itex]\sin{2x}=\sin{\pi}=0[/itex].

The 2x as opposed to just x makes a difference to where the min and max are located. [itex]\sin(ax)[/itex] for some constant a will either squish or expand the sine graph.
 
  • #13
sin(2x)=1

arcsin(sin(2x))=arcsin(1)

x=arcsin(1)/2

45°.. sin(90°-x) = sin (90°+x)

90-45 = 45 ?
 
  • #14
xtrubambinoxpr said:
sin(2x)=1

arcsin(sin(2x))=arcsin(1)

x=arcsin(1)/2

45°
Up to this point, the maths tells you that the maximum is located at [itex]x=45^o[/itex], right?

xtrubambinoxpr said:
sin(90°-x) = sin (90°+x)

90-45 = 45 ?

That equality is only valid for the graph [itex]y=\sin{x}[/itex] since the line of symmetry (the line that passes through the maximum) is located at [itex]x=90^o[/itex]. Now we're dealing with [itex]y=\sin(2x)[/itex] where you've found that the maximum is located at [itex]x=45^o[/itex] so in this case, for [itex]y=\sin(2x)[/itex], we have that [itex]\sin(45^o+x)=\sin(45^o-x)[/itex].

So if 39.3o is one solution, then what's the other?

It would also help if you drew the graphs of sin(x) and sin(2x).
 
  • #15
Mentallic said:
Up to this point, the maths tells you that the maximum is located at [itex]x=45^o[/itex], right?



That equality is only valid for the graph [itex]y=\sin{x}[/itex] since the line of symmetry (the line that passes through the maximum) is located at [itex]x=90^o[/itex]. Now we're dealing with [itex]y=\sin(2x)[/itex] where you've found that the maximum is located at [itex]x=45^o[/itex] so in this case, for [itex]y=\sin(2x)[/itex], we have that [itex]\sin(45^o+x)=\sin(45^o-x)[/itex].

So if 39.3o is one solution, then what's the other?

It would also help if you drew the graphs of sin(x) and sin(2x).

ahhh i see so the first value i get for x will either be added or subtracted from the half period or however you want to phrase it. so if its 39.3° then 90°-39.3° = 50.7°
 
  • #16
xtrubambinoxpr said:
ahhh i see so the first value i get for x will either be added or subtracted from the half period or however you want to phrase it. so if its 39.3° then 90°-39.3° = 50.7°

Yep, exactly! For future reference, you should always draw your graphs, label any important info, and try to deduce any facts you may know about the symmetry or special features of the graph. It will help you immensely.
 
  • #17
Mentallic said:
Yep, exactly! For future reference, you should always draw your graphs, label any important info, and try to deduce any facts you may know about the symmetry or special features of the graph. It will help you immensely.

thanks man! really appreciate it!
 
  • #18
Happy to help :smile:
 

Related to Trig Problem: Where Did 50 Degrees Come From in Sin(2x)=.98?

What is the basic definition of Trigonometry?

Trigonometry is a branch of mathematics that deals with the study of triangles and the relationships between their sides and angles.

What is the sine function in Trigonometry?

The sine function is one of the primary trigonometric functions and is defined as the ratio of the opposite side of a right triangle to its hypotenuse.

How do you solve a Trigonometry problem involving the sine function?

To solve a Trigonometry problem involving the sine function, you can use the basic trigonometric identities, such as the Pythagorean identity or the double angle identity, along with algebraic manipulations to simplify the expression and find the value of the sine function.

What is the double angle identity for the sine function?

The double angle identity for the sine function is sin(2x) = 2sin(x)cos(x). This identity is useful for simplifying expressions and solving Trigonometry problems involving the sine function.

Can you provide an example of a Trigonometry problem involving the sine function?

Yes, for example, if we have the problem sin(2x) = 1/2, we can use the double angle identity to rewrite it as 2sin(x)cos(x) = 1/2. Then, by setting each factor equal to 1/2, we can solve for x and find that x = π/12 or 5π/12.

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